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# 11.9: Solving Equations Using the Pythagorean Theorem

Difficulty Level: At Grade Created by: CK-12
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Practice Solving Equations Using the Pythagorean Theorem
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What if you knew the lengths of two sides of a right triangle but not the third? How could you find the length of this missing side? After completing this Concept, you'll be able to use the Pythagorean Theorem to solve problems like this one where variables are involved.

### Guidance

In the previous concept, we learned about the Pythagorean theorem and how to use it to find the hypotenuse. In this concept, we will learn how to use the Pythagorean theorem to find any side of a right triangle.

#### Example A

Determine the value of the missing side. You may assume that each triangle is a right triangle.

Solution

Apply the Pythagorean Theorem.

$a^2+b^2 &= c^2\\x^2+15^2 &= 21^2\\x^2+225 &= 441\\x^2 &= 216 \Rightarrow \\x & =\sqrt{216}=6 \sqrt{6}$

#### Example B

Determine the values of the missing sides. You may assume that each triangle is a right triangle.

Solution

Apply the Pythagorean Theorem.

$a^2+b^2 &= c^2\\18^2+15^2 &= z^2\\324+225 &= z\\z^2 &= 549 \Rightarrow \\z & =\sqrt{549}=3 \sqrt{61}$

#### Example C

One leg of a right triangle is 5 units longer than the other leg. The hypotenuse is one unit longer than twice the size of the short leg. Find the dimensions of the triangle.

Solution

Let $x =$ length of the short leg.

Then $x + 5 =$ length of the long leg

And $2x + 1 =$ length of the hypotenuse.

The sides of the triangle must satisfy the Pythagorean Theorem.

$\text{Therefore:} && x^2+(x+5)^2& =(2x+1)^2\\\text{Eliminate the parentheses:} && x^2+x^2+10x+25& =4x^2+4x+1\\\text{Move all terms to the right hand side of the equation:} && 0& =2x^2-6x-24\\\text{Divide all terms by} \ 2: && 0& =x^2-3x-12\\\text{Solve using the quadratic formula:} && x& =\frac{3 \pm \sqrt{9+48}}{2}=\frac{3 \pm \sqrt{57}}{2}\\&& x& =\underline{\underline{5.27}} \ \text{or} \ x=-2.27$

The negative solution doesn’t make sense when we are looking for a physical distance, so we can discard it. Using the positive solution, we get: $\text{short leg} = 5.27, \text{long leg} = 10.27$ and $\text{hypotenuse} = 11.54$ .

Watch this video for help with the Examples above.

### Vocabulary

• The Pythagorean Theorem is a statement of how the lengths of the sides of a right triangle are related to each other. A right triangle is one that contains a 90 degree angle. The side of the triangle opposite the 90 degree angle is called the hypotenuse and the sides of the triangle adjacent to the 90 degree angle are called the legs .

• If we let $a$ and $b$ represent the legs of the right triangle and $c$ represent the hypotenuse then the Pythagorean Theorem can be stated as:

In a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs. That is: $a^2+b^2=c^2$ .

### Guided Practice

Determine the values of the missing sides. You may assume that each triangle is a right triangle.

Solution

Apply the Pythagorean Theorem. $a^2+b^2 &= c^2\\y^2+3^2 &= 7^2\\y^2+9 &= 49\\y^2 &= 40 \Rightarrow\\y & =\sqrt{40}=2 \sqrt{10}$

### Practice

Find the missing length of each right triangle.

1. $a = 12, b = 16, c = ?$
2. $a = ?, b = 20, c = 30$
3. $a = 4, b = ?, c = 11$
4. $a = 12, b = ?, c = 37$
5. One leg of a right triangle is 4 feet less than the hypotenuse. The other leg is 12 feet. Find the lengths of the three sides of the triangle.
6. One leg of a right triangle is 3 more than twice the length of the other. The hypotenuse is 3 times the length of the short leg. Find the lengths of the three legs of the triangle.
7. Two sides of a right triangle are 5 units and 8 units respectively. Those sides could be the legs, or they could be one leg and the hypotenuse. What are the possible lengths of the third side?

Oct 01, 2012

Sep 15, 2014